How to use the Chain rule for: $y=\cosh(a\sinh^{-1}x)$

Question

$$y=\cosh(a\sinh^{-1}x)$$ where $a$ is a constant. What do i substitute for $u$ and $v$ to then find $\frac{du}{dx}$ and $\frac{dv}{dx}$?

I am then suposed to prove that: $$(x^2+1)\frac{d^2y}{dx^2} +x\frac{dy}{dx}-a^2y=0$$

Answer 1

HINT:

So, $$\cosh^{-1}y=a\sinh ^{-1}x$$

Differentiating wrt $x,$ $$\frac1{\sqrt{y^2-1}}\frac{dy}{dx}=\frac a{\sqrt{1+x^2}}$$

Squaring we get, $$(1+x^2)\left(\frac{dy}{dx}\right)^2=a^2(y^2-1)$$

Again differentiating wrt $x,$

 $$2x\left(\frac{dy}{dx}\right)^2+(1+x^2)2\frac{dy}{dx}\frac{d^2y}{dx^2}=a^2\cdot2y\frac{dy}{dx}$$

Divide by $2\frac{dy}{dx}$ assuming $\frac{dy}{dx}\ne0$